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Standard XII

Chemistry

Gay Lussac's Law, Avagadro's Law

A two litre v...

Question

A two litre vessel contains 0.48 mole of $C O_{2}$ , 0.48 mole of $H_{2}$ , 0.96 mole of $C O$ at equilibrium.
$C O_{2} (g) + H_{2} (g) ⇌ H_{2} O (g) + C O (g)$
How many $g m$ (only integer part) of $H_{2}$ must be added to bring the concentration of $C O$ to $0.6 M$ ?

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Solution

We Know that According to Lechatlier Principle, when we increase the concentration of reactant at the equilibrium shifts in the forward direction.
in this question,we adding H2 which is a reactant and concentration of CO is decreasing which is product which is not possible.

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Similar questions

Q. The equilibrium constant for the reaction

C O (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g)

is 5. How many moles of

C O_{2}

must be added to 1 litre container already containing 3 moles each of CO and

H_{2} O

to make 2 M equilibrium concentration of CO?

Q. The equilibrium constant for,

H_{2} (g) + C O_{2} (g) ⇌ H_{2} O (g) + C O (g)

1.80

1000^{o}

C. If

1.0

mole of

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and

1.0

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are placed in one litre flask, the final equilibrium concentration of

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C will be:

Q. An equilibrium mixture of

C O (g) + H_{2} O (g) ⇌ {C O}_{2} (g) + H_{2}

present in a vessel of one litre capacity at

815^{0}

C was found by analysis to contain 0.4 mol of

C O

, 0.3 mol of

H_{2} O

, 0.2 mol of

{C O}_{2}

and 0.6 mol of

H_{2}

a) Calculate

K_{C}

.
b) If it is derived to increase the concentration of

C O

to 0.6 mol by adding

{C O}_{2}

to the vessel, how many moles must be added into equilibrium mixture at a constant temperature in order to get this change?

Q. For the reaction,

{C O}_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)

the value of Kc at

552^{\circ} C

is 0.137. If 5 moles of

{C O}_{2}

, 5 moles of

H_{2}

, 1 mole of

C O

and 1 mole of

H_{2} O

are initially present what are the actual concentrations of

{C O}_{2}, H_{2}, C O, H_{2} O

at equilibrium?

Q. An equilibrium mixture of

C O (g) + H_{2} O (g) ⇋ C O_{2} (g) + H_{2} (g)

, present in a vessel of one litre capacity at

815^{\circ} C

was found by analysis to contain 0.4 mole of CO, 0.3 mole of

H_{2} O

, 0.2 mole of

C O_{2}

and 0.6 mole of

H_{2}

(a) Calculate

K_{c}

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(b) If it is derived to increase the concentration of CO to 0.6 moles by adding

C O_{2}

to the vessel, how many moles must be added into equilibrium mixture at a constant temperature in order to get this change?

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A two litre vessel contains 0.48 mole of CO2, 0.48 mole of H2, 0.96 mole of CO at equilibrium. CO2(g)+H2(g)⇌H2O(g)+CO(g) How many gm (only integer part) of H2 must be added to bring the concentration of CO to 0.6M?

We Know that According to Lechatlier Principle, when we increase the concentration of reactant at the equilibrium shifts in the forward direction.in this question,we adding H2 which is a reactant and concentration of CO is decreasing which is product which is not possible.

A two litre vessel contains 0.48 mole of $C O_{2}$ , 0.48 mole of $H_{2}$ , 0.96 mole of $C O$ at equilibrium.
$C O_{2} (g) + H_{2} (g) ⇌ H_{2} O (g) + C O (g)$
How many $g m$ (only integer part) of $H_{2}$ must be added to bring the concentration of $C O$ to $0.6 M$ ?

We Know that According to Lechatlier Principle, when we increase the concentration of reactant at the equilibrium shifts in the forward direction.
in this question,we adding H2 which is a reactant and concentration of CO is decreasing which is product which is not possible.